Question

Dry air was bubbled through a bulb containing $22g$ of an non-volatile organic compound in $150g$ of water, then it passed through a bulb containing water at the same temperature and finally through a bulb containing anhydrous calcium chloride. The weight loss in bulb containing water was $0.075g$
and gain in mass of the calcium chloride tube was $1.5g$.
Calculate the molecular mass of the organic substance.

• A. $42.16g/mol$
• B. $50.16g/mol$
• C. $59.16g/mol$
• D. $64.16g/mol$

Answer 1

The correct option is B $50.16g/mol$

Given,
$\text{Gain in mass of the}CaC{l}_2=1.5g$
$\text{Loss in mass of the pure solvent}=0.075g$
By Ostwald-Walker method,
We have,

$\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{\text{Loss in mass of solvent bulb}}{\text{Gain in mass of}CaC{l}_2\text{tube}}$
$\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{0.075}{1.5}$
For a non-volatile solute,
$p_s=p^{\circ }\times X_{solvent}\therefore \frac{p^{\circ }-p^{\circ }.X_{solvent}}{p^{\circ }}=\frac{0.075}{1.5}\frac{1-X_{solvent}}{1}=\frac{0.075}{1.5}\frac{X_{solute}}{1}=\frac{0.075}{1.5}\frac{n_{solute}}{n_{total}}=\frac{0.075}{1.5}$
where,
$n_{solute}$is the number of moles of solute
$n_{total}$is the number of moles of solute and solvent.
Let the molecular mass of the organic substance be 'm'
thus,
$\frac{n_{solute}}{n_{solute}+n_{solvent}}=\frac{w/m}{\frac{w}{m}+\frac{W}{M}}$
Where,
M is molar mass of water
W is mass of solvent
w is mass of solute
m is molar mass of solute
$\therefore$
$\frac{0.075}{1.5}=\frac{\frac{22}{m}}{\frac{22}{m}+\frac{150}{18}}\Rightarrow \frac{22}{22+\frac{150}{18}m}\Rightarrow \frac{0.075}{1.5}\times [22+\frac{150}{18}m]=22\Rightarrow m=50.16g/mol$